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If $A,B$ are non-unital commutative algebras over a field $R$, what should their (categorical) coproduct?

I know for unital algebras the tensor product is the coproduct, but I think the construction of tensor product as a coproduct only worked there because we could have a unit element in multiplication, so we could just send, for example, $a\otimes 1_B \to f(a)g(1_B)=f(a)$ for some morphisms $f:A\to C$, $g:B\to C$ where $C$ is another algebra.

Apparently without unit element we can't do this anymore. So what would coproduct be in this case?

Smiley1000
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user160738
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  • Do you assume $A$ and $B$ to be associative? – Smiley1000 Dec 18 '24 at 18:37
  • Related questions: https://math.stackexchange.com/q/197819/491450 , https://math.stackexchange.com/q/143098/491450 , https://math.stackexchange.com/q/625874/491450 , https://math.stackexchange.com/q/1972781/491450 , https://math.stackexchange.com/q/3865235/491450 – Smiley1000 Dec 18 '24 at 19:15

3 Answers3

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The coproduct of $A$ and $B$ has underlying $R$-vector space $A \oplus B \oplus (A\otimes B)$ and with multiplication defined by bilinearity and by $$(a_1,b_1,a'_1\otimes b'_1) \cdot (a_2,b_2,a'_2\otimes b'_2)= (a_1\cdot a_2,b_1\cdot b_2,a_1\otimes b_2 + a_2\otimes b_1 + (a_1\cdot a'_2)\otimes b'_2+(a'_1\cdot a_2)\otimes b'_1 + a'_2\otimes(b_1\cdot b'_2) + a'_1\otimes (b'_1\cdot b_2) + (a'_1\cdot a'_2)\otimes (b'_1\cdot b'_2))$$ for $a_1,a_2,a'_1,a'_2$ in $A$ and $b_1,b_2,b'_1,b'_2$ in $B$. The the first coproduct inclusion sends $a$ to $(a,0,0)$ and the second sends $b$ to $(0,b,0)$.

Nex
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  • As a remark, one can notice that $(a_1,b_1,a'_1\otimes b'_1)$ is essentially the "sum of $3$ things", $(a_2,b_2,a'_2\otimes b'_2)$ is essentially the "sum of $3$ things" and $(a_1\cdot a_2,b_1\cdot b_2,a_1\otimes b_2 + a_2\otimes b_1 + (a_1\cdot a'_2)\otimes b'_2+(a'_1\cdot a_2)\otimes b'_1 + a'_2\otimes(b_1\cdot b'_2) + a'_1\otimes (b'_1\cdot b_2) + (a'_1\cdot a'_2)\otimes (b'_1\cdot b'_2))$ is essentially the "sum of $9$ things". This may give us confidence that we have expanded everything out correctly. – Smiley1000 Dec 18 '24 at 18:51
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Just to provide a motivation for @Nex's answer, notice that the forgetful functor $F \colon \textbf{UnitalCommAlg}_k \Rightarrow \textbf{NonUnitalCommAlg}_k$ has a left-adjoint functor $(-)^+ \colon \textbf{NonUnitalCommAlg}_k \Rightarrow \textbf{UnitalCommAlg}_k$, known as the unitalisation functor, given by $A \mapsto A \oplus k$ and $(a_1, r_1)(a_2, r_2) = (a_1a_2 + a_1r_2 + a_2r_1, r_1r_2)$. Intuitively, if you are given a $R$-algebra without a unit, then you would like to "force" a unit onto it by taking $\operatorname{span}_k(A \cup \{1_k\})$. (I like to think of it as analogous to when you consider the vector space $V \oplus k$ to represent affine transformations and not just linear maps.) And since left-adjoints preserve colimits, $(A \coprod B)^+ = A^+ \coprod B^+$, and expanding gives $(A \coprod B) \oplus k = (A \oplus k) \coprod (B \oplus k) = (A \otimes B) \oplus A \oplus B \oplus k$, where we used the fact that coproducts in $\textbf{UnitalCommAlg}_k$ is tensor product over $k$.

ketsi
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Let $R$ be a field. Consider the category $R\mathsf{-NonUnitalAssocCommAlg}$ of non-unital associative commutative $R$-algebras.

Here is a way to construct the coproduct of a family of non-unital associative commutative $R$-algebras $A_i$ in the category $R\mathsf{-NonUnitalAssocCommAlg}$:

Let $B = \operatorname{Free}(\{x_{i, a} : i \in I, a \in A_i\})$ be the free non-unital associative commutative $R$-algebra on the indeterminates $x_{i, a}$ for $i \in I$ and $a \in A_i$. Now let $J \subseteq B$ be the ideal generated by expressions of the following form:

  • $x_{i, a_1} + x_{i, a_2} - x_{i, a_1 + a_2}$ for $i \in I$ and $a_1, a_2 \in A_i$
  • $x_{i, a_1} x_{i, a_2} - x_{i, a_1 a_2}$ for $i \in I$ and $a_1, a_2 \in A_i$
  • $x_{i, r} - r$ for $i \in I$ and $r \in R$

Now take $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} C = \quotient{B}{J}$. Then $C \cong \coprod_{i \in I} A_i$ is the coproduct over all the $A_i$ in $R\mathsf{-NonUnitalAssocCommAlg}$. The inclusions $A_i \to C$ are the obvious ones, and the fact that we took the quotient by $J$ ensures that these are actually $R$-algebra homomorphisms.

Here are some related questions:

Smiley1000
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